2 Answers
2

Take $b=2$ and $x=8$, in which case $x=1\cdot 2^3+0\cdot 2^2+0\cdot 2+0\cdot 1$, i.e., $a_0=a_1=a_2=0$ and $a_3=1$. Clearly the condition $a_3\ge b/4$ is satisfied since $a_3=1> b/4=1/2$ and $\lfloor\sqrt{x}\rfloor=2$. On the other hand, if $a_0$ becomes equal to $1$, then $x$ becomes $9$ and $\lfloor\sqrt{9}\rfloor=3$.

Also try $x=3024$, with $b=10$, where $a_3=3>10/4$, $\lfloor\sqrt{x}\rfloor=54$ and $\lfloor\sqrt{x+1}\rfloor=55$.